Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in JAVA):

Solution:
import java.io.DataInputStream; 
import java.io.FileInputStream; 
import java.io.IOException; 
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner; 
import java.util.StringTokenizer; 

public class Main 
{ 
	static final long mod = (int)1e9+7;

	static class Reader 
	{ 
		final private int BUFFER_SIZE = 1 << 16; 
		private DataInputStream din; 
		private byte[] buffer; 
		private int bufferPointer, bytesRead; 

		public Reader() 
		{ 
			din = new DataInputStream(System.in); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public Reader(String file_name) throws IOException 
		{ 
			din = new DataInputStream(new FileInputStream(file_name)); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public String readLine() throws IOException 
		{ 
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c; 
			while ((c = read()) != -1) 
			{ 
				if (c == '\n') 
					break; 
				buf[cnt++] = (byte) c; 
			} 
			return new String(buf, 0, cnt); 
		} 

		public int nextInt() throws IOException 
		{ 
			int ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do
			{ 
				ret = ret * 10 + c - '0'; 
			} while ((c = read()) >= '0' && c <= '9'); 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		public long nextLong() throws IOException 
		{ 
			long ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
			if (neg) 
				return -ret; 
			return ret; 
		} 

		public double nextDouble() throws IOException 
		{ 
			double ret = 0, div = 1; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 

			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 

			if (c == '.') 
			{ 
				while ((c = read()) >= '0' && c <= '9') 
				{ 
					ret += (c - '0') / (div *= 10); 
				} 
			} 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		private void fillBuffer() throws IOException 
		{ 
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
			if (bytesRead == -1) 
				buffer[0] = -1; 
		} 

		private byte read() throws IOException 
		{ 
			if (bufferPointer == bytesRead) 
				fillBuffer(); 
			return buffer[bufferPointer++]; 
		} 

		public void close() throws IOException 
		{ 
			if (din == null) 
				return; 
			din.close(); 
		} 
	} 
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	}
	
	public static void main(String[] args) throws IOException 
	{ 
		Reader sc=new Reader();
		int t = sc.nextInt(); 
		while (t-- > 0) 
		{ 
			int l = sc.nextInt(); 
			int r = sc.nextInt(); 
			int d = sc.nextInt(); 
			int u = sc.nextInt(); 

			int x = sc.nextInt(); 
			int y = sc.nextInt(); 
			int xl = sc.nextInt(); 
			int yd = sc.nextInt(); 
			int xr = sc.nextInt(); 
			int yu = sc.nextInt(); 

			int ans = 1;
			
			if(r-l > xr-x || l-r > x-xl)
				ans = 0;
			
			if(u-d > yu-y || d-u > y-yd)
				ans = 0;
			if(xl == xr && (l+r) > 0)
				ans = 0;
			if(yd == yu && (d+u) > 0)
				ans = 0;
			
			if(ans == 1)
			{
				System.out.println("YES");
			}
			else {
				System.out.println("NO");
			}
		} 				

	} 
	
	
}